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OPERATIONS RESEARCH
Vol. 63, No. 3, May–June 2015, pp. 512–526
ISSN 0030-364X (print) — ISSN 1526-5463 (online)
http://dx.doi.org/10.1287/opre.2015.1373
© 2015 INFORMS
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Error Theory for Elimination by Aspects
Rajeev Kohli, Kamel Jedidi
Graduate School of Business, Columbia University, New York, New York 10027
{[email protected], [email protected]}
Elimination by aspects (EBA) is a random utility model that is considered to represent the choice process used by consumers
more faithfully than logit and probit models. One limitation of the model is that it does not have a known error theory. We
show that EBA can be derived by assuming that aspects have random utilities with independent, extreme value distributions.
Multinomial logit and rank-ordered logit models are special cases of EBA.
Subject classifications: elimination by aspects; Luce axiom; multinomial logit model; rank-ordered logit model;
independence from irrelevant alternatives; lexicographic choice; choice models.
Area of review: Marketing Science.
History: Received June 2014; revisions received September 2014, December 2014, January 2015; accepted February
2015. Published online in Articles in Advance May 13, 2015.
1. Introduction
considered the ability to accommodate violations of order
independence to be a significant advantage of EBA over
other choice models. McFadden (1981) endorsed the view
and remarked that EBA had significant potential for econometric applications because it allowed complex patterns of
substitutability among alternatives. Still, EBA has seen little practical use. One possible reason is that it does not
have an error theory. Another is that it appears to be too far
removed from the utility maximization paradigm that has
come to dominate the literature on empirical choice models
over the last half century.
The objectives of this paper are to (i) derive an error
theory for EBA, (ii) examine its relation with choice models that assume utility maximization, and (iii) consider its
implications for model estimation. We discuss each objective below.
(i) We show that EBA can be obtained when each aspect
has random utility with an independent, extreme value distribution. We develop the theory in three steps. The first
step considers a lexicographic choice rule that uses aspects
arranged in decreasing order of their utilities. If each aspect
has random utility with extreme value distribution, then the
probability of using a particular aspect ordering in a lexicographic rule is specified by the rank-ordered logit model
(Beggs et al. 1981). The second step partitions the set of all
possible aspect orderings into mutually exclusive and collectively exhaustive subsets. Aspect orderings in each such
subset are equivalent in the following sense: when used in
a lexicographic rule, they each eliminate alternatives in any
choice set in the same sequence, using the same subset of
aspects in the same order. The third step shows that there
is a one-to-one mapping between the probabilities associated with each subset of equivalent aspect orderings and
the EBA choice probabilities.
Elimination by aspects (EBA) is a theory of choice that
was proposed by Tversky (1972a, b) more than 40 years
ago. It is considered to reflect the choice process used by
consumers more faithfully than traditional choice models.
EBA does not assume that a consumer chooses an alternative to maximize utility. Instead, it views choice to be
the result of a probabilistic process that eliminates alternatives in stages. Tversky (1972a, p. 281) described EBA as
follows:
In this theory, each alternative is viewed as a set of aspects.
At each stage in the process, an aspect is selected (with probability proportional to its weight), and all the alternatives that
do not include the selected aspect are eliminated. The process
continues until all alternatives but one are eliminated.
A key feature of EBA is that it does not assume order
independence, a condition that is probably the weakest
form of independence from irrelevant alternatives (IIA).
Unlike (strong) IIA, order independence only requires that
the ordering—not necessarily the ratio—of the choice probabilities for two alternatives should be independent of any
other alternatives in a choice set.1 But as the following classic example by Debreu (1960) illustrates, even this seemingly mild condition can be violated. Consider a choice
set with two alternatives, A and B, and a person who
chooses A with a higher probability than B. Now suppose
one or more alternatives that are essentially identical to
A are added to the choice set. Then order independence
is violated if their addition has no effect on the choice
probability for B but lowers the choice probability of A
to less than 1/2. Experiments demonstrating violations of
order independence were reported in the 1960s by Becker
et al. (1963), Chipman (1960), Coombs (1958), Krantz
(1967) and Tversky and Russo (1969). Tversky (1972a, b)
512
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Operations Research 63(3), pp. 512–526, © 2015 INFORMS
(ii) The error theory implies that EBA is a type of
logit model. It replaces the assumption that a consumer
maximizes utility when selecting an alternative by the assumption that a consumer maximizes utility when selecting
the next aspect to use for eliminating alternatives. In other
words, EBA changes the choice process used by consumers
(from trade-offs among aspects to sequential elimination
by aspects), but it is consistent with utility maximization
when considering the order in which aspects are used to
eliminate alternatives. An EBA choice probability reduces
to the probability obtained from a rank-ordered logit model
when EBA selects an alternative using as many stages as
there are aspects. It reduces to a multinomial logit choice
probability if we define each alternative to have a unique
aspect.
(iii) Our analysis implies that the parameters of an EBA
model may be estimated by modifying maximum likelihood
procedures that are used for estimating logit models. Maximum likelihood estimates have well-known properties, and
their statistical significance can be assessed using parametric methods. It also implies that EBA can be extended in
ways analogous to standard random utility models. Examples include allowing the deterministic aspect utilities to
be functions of covariates and developing latent class and
random-effects models to reflect consumer heterogeneity in
aspect utilities.
Background. Although Tversky did not obtain an error
model for EBA, he showed that it satisfies the conditions for
a random utility model. These conditions were described by
Block and Marschak (1960) and require that a random utility model must (i) specify a probability distribution over the
set of ordered sequences of alternatives in a choice set, and
(ii) choose an alternative from a choice set with a probability that is obtained by adding the probabilities of occurrence
of those sequences in which it has the highest rank. Tversky
considered associating random utilities with aspects but
concluded that this would not lead to a characterization of
EBA by an additive utility function over aspects (Tversky
and Sattah 1979). McFadden (1981, p. 226) observed:
The EBA functional form has considerable potential for
econometric applications. When the scale functions V are
log-linear in parameters, ln V 4zA 5 = ‚0A zA , the choice probabilities can be written as sums of products of MNL forms.
Maximum likelihood estimation could be carried out for
such systems with relatively minor modifications of current
MNL computer programs. One drawback of EBA for econometric applications is that the motivation for the model provides little guidance for the parametric specification of the
scale function V .
Even as he noted that the Luce choice rule for selecting
an aspect could be represented by the multinomial logit formula, McFadden refrained from proposing that the aspects
be considered to have independent random utilities with
extreme value distributions. The EBA choice probabilities
513
would then have followed trivially upon the further assumption that aspect choices are independent across stages—
that is, by assuming that an aspect’s utility changes as a
person moves from one stage of elimination to the next. But
this assumption is inappropriate. It implies, for example,
that an aspect (e.g., low price) could have lowest utility at
one stage of elimination and the highest utility at the next
stage of elimination. Such an assumption is inconsistent
with Tversky’s view, who stated (Tversky 1972a, p. 296):
The EBA model accounts for choice in terms of a covert
elimination process based on sequential selection of aspects.
Any such sequence of aspects can be regarded as a particular
state of mind which leads to a unique choice. In light of this
interpretation, the choice mechanism at any given moment in
time is entirely deterministic; the probabilities merely reflect
the fact that at different moments in time different states of
mind (leading to different choices) may prevail.
In other words, Tversky’s view was that the ordering of
aspects is fixed on a given choice occasion but can change
across choice occasions.2 The error model we consider is
consistent with this view. We show that an extreme value
distribution for aspect utilities is appropriate even when
aspect choices are not independent across stages.
Although no previous research has considered an error
theory for EBA, researchers have (i) proposed models that
are related to EBA (Corbin and Marley 1974, Manrai and
Sinha 1989, Pihlens 2008), (ii) considered the relation
between nested logit models and EBA (Batley and Daly
2006), (iii) examined the use of EBA for modeling consumer response to promotions (Fader and McAlister 1990),
(iv) proposed nonparametric estimation procedures (Görür
et al. 2006, Gilbride and Allenby 2006), and (v) discussed
issues concerning the degrees of freedom in EBA models
(Batsell et al. 2003, Park and Choi 2013).
Specifically, Corbin and Marley (1974) introduced a
random utility model that generalizes EBA by relaxing
the assumption that choice probabilities satisfy the multiplicative inequality. Manrai and Sinha (1989) proposed
elimination-by-cutoffs, a model that considers choice to be
the result of a consumer choosing cutoffs in a multidimensional perceptual space. Pihlens (2008) proposed a multiattribute elimination by aspects (MEBA) model in which
the alternatives are described by attributes, and the aspects
correspond to main effects, and selected interaction effects,
among the attributes. Batley and Daly (2006) considered a
special case of EBA with two aspects and three alternatives
and showed that, in this case, there is a one-to-one mapping between the EBA parameters and the parameters of a
nested logit model. Görür et al. (2006) used pairwise comparisons of alternatives, and Gilbride and Allenby (2006)
used a nonparametric Bayesian approach, to estimate EBA
models. Batsell et al. (2003) and Park and Choi (2013)
examined issues concerning degrees of freedom in the estimation of EBA. The former also showed that there is a
linear relationship between the EBA parameters and the
differences in choice probabilities across choice sets.
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Operations Research 63(3), pp. 512–526, © 2015 INFORMS
Organization of the paper. Section 2 describes EBA and
introduces a probabilistic lexicographic rule obtained when
aspects have independent, extreme value distributions. Section 3 shows that EBA is equivalent to the probabilistic
lexicographic rule. The equivalence establishes the error
structure for EBA. Section 4 describes a probabilistic utility
function for EBA. Section 5 discusses the implications for
estimating EBA parameters. Section 6 concludes the paper.
the precedence conditions also imply that 4kj 5 < 4k5
for any aspect k ∈ Aj that is distinct from kj+1 1 0 0 0 1 km and
is not used to eliminate alternatives at any step, because
(i) either EBA terminates before using all remaining aspects
in Aj , or (ii) aspect k appears in all or none of the alternatives in at least one of the sets Cj 1 0 0 0 1 Cm .
The following algorithm gives a formal description of
EBA.
2. EBA and Probabilistic Lexicographic
Rule
Initialization step: Let A0 = A = 811 0 0 0 1 n9, C0 = C,
xi = 4xi1 1 0 0 0 1 xin 5, for all i ∈ C0 , and  = 41 1 0 0 0 1 n 5.
Iteration step j 4¾15: Let
We describe EBA and a probabilistic lexicographic rule.
Let C denote a choice set. Let A = 811 0 0 0 1 n9 denote the
set of n aspects that are used to describe the alternatives
in C. An aspect can be a nominal attribute (e.g., color) or a
threshold value associated with a continuous attribute (e.g.,
a price cutoff). Let xik = 1 if alternative i ∈ C has aspect
k ∈ A; otherwise, xik = 0. Let xi = 4xi1 1 0 0 0 1 xin 5 denote the
profile of alternative i ∈ C. We assume that all alternatives
in C are distinct; that is, xi 6= xi0 for any two alternatives
i1 i0 ∈ C.
2.1. Elimination by Aspects
We begin with an informal description of EBA and then
provide a more precise description.
EBA assumes that a person associates an importance
weight k > 0 with each aspect k ∈ A. It uses an iterative
procedure to eliminate alternatives from the choice set C.
The first stage (step) identifies a subset of aspects A1 ⊆ A
that appear in at least one, but not all, alternatives in C.
An aspect k1 ∈ A1 is selected with probability proportional
to k1 . Alternatives in which aspect k1 does not appear
(that is, for which xik1 = 0) are eliminated. Let C1 ⊂ C
denote the set of remaining alternatives. The elimination
of alternatives stops if C1 has only one alternative. Otherwise, the procedure advances to a second stage, where
it identifies a subset of aspects A2 ⊂ A1 that appear in at
least one, but not all, alternatives in C1 . An aspect k2 ∈ A2
is selected with probability proportional to k2 . Alternatives in which aspect k2 does not appear (that is, for which
xik2 = 0) are eliminated. The procedure iterates until all
alternatives except one are eliminated.
We will need the following (precedence) conditions to
show that EBA is equivalent to the probabilistic lexicographic rule discussed in §2.2. For each step j, define
4kj 5 < 4k5 to mean that aspect kj is selected for eliminating alternatives before any other aspect k 6= kj , k ∈ Aj .
For example, if A = 811 21 39 and k1 = 1 is selected at step
j = 1, then 415 < 425 and 415 < 435. These conditions mean that aspect 1 is used to eliminate alternatives
before aspects 2 and 3 (which may or may not be used
in a later elimination step). If there are m ¶ n elimination
steps, then 4k1 5 < 4k2 5 < · · · < 4km 5, because aspects
that are used to eliminate alternatives at a later step are also
available for eliminating alternatives at earlier steps. But
X
Aj = k ∈ Aj−1 0 <
xik < —Cj−1 —
i∈Cj−1
denote the subset of aspects in Aj−1 that appear in at least
one, but not all, of the alternatives in Cj−1 . Select aspect
kj ∈ Aj with probability
q4kj 1 Aj 5 = P
kj
k∈Aj
k
0
(1)
Set Cj = 8i ∈ Cj−1 — xikj = 19. Let 4kj 5 < 4k5, for all
k ∈ Aj \8kj 9, to indicate that aspect kj is used to eliminate
alternatives before any other aspect in Aj .
Termination step: Stop if —Cj — = 1.
Since each step eliminates at least one alternative, EBA
terminates in m ¶ n steps. We refer to a particular sequence
k1 1 0 0 0 1 km of aspects used by EBA as an instance. Observe
that Cj ≡ Cj 4kj 5—that is, Cj is a function of kj —because
its composition depends on which aspect kj ∈ Aj is selected
to eliminate alternatives from Cj−1 . Unless necessary, we
simplify notation by writing Cj instead of the more explicit
Cj 4kj 5 in the rest of the paper.
Tversky (1972a) noted that the EBA choice probability
for an alternative can be computed by using the following recursive formula. Let Aij ⊆ Aj denote the subset of
aspects in Aj that appear in alternative i ∈ Cj−1 . As defined
previously, let Cj 4kj 5 denote the subset of alternatives in
Cj−1 that have aspect kj . Let p4i1 Cj 4kj 55 denote the choice
probability for alternative i, conditional on the selection of
aspect kj ∈ Aij at step j. Then the probability that EBA
chooses alternative i ∈ Cj−1 is given by
p4i1 Cj−1 5 =
X
kj
P
kj ∈Aij
k∈Aj
k
p4i1 Cj 4kj 551
j ¾ 11
P
where kj / k∈Aj k is the probability of selecting aspect
kj ∈ Aj at step j. The unconditional choice probability for
alternative i ∈ C is p4i1 C0 5, obtained by setting C = C0 .
The following example describes an EBA problem, illustrates the precedence conditions, computes the choice probabilities for alternatives, and shows how EBA allows violations of order independence.
Kohli and Jedidi: EBA Error Theory
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Operations Research 63(3), pp. 512–526, © 2015 INFORMS
Example 1 (Beethoven and Debussy Record Albums).
Consider the following choice problem due to Debreu
(1960). A person has to choose an alternative from the set
C = 8x1 1 x2 1 x3 9. Alternatives x1 and x2 are different but
equally good recordings of the same Beethoven symphony.
Alternative x3 is a recording of a Debussy suite.3 Let
A = 811 21 31 49 denote the set of four aspects that are
used to describe the albums. Aspect 1 is “Beethoven
symphony,” aspect 2 is “Debussy suite,” and aspects 3
and 4 are features that distinguish between the two
Beethoven albums, such as two different orchestras or
recording labels. Let k denote the importance weight that
the person associates with aspect k, for each k = 11 21 31 4.
Let xi = 4xi1 1 0 0 0 1 xi4 5 denote the profile of alternative i,
where xik = 1405 if alternative i has (does not have)
aspect k. Then x1 = 41 0 1 05 and x2 = 41 0 0 15 represent
the two Beethoven albums, and x3 = 40 1 0 05 represents
the Debussy album.
We describe the steps for one instance of EBA in detail
and then give a summary description of the other instances.
Initialization step. Set A0 = A = 811 21 31 49, C0 = C =
811 21 39, x1 = 41 0 1 051 x2 = 41 0 0 151 x3 = 40 1 0 05
and  = 41 1 0 0 0 1 4 5.
Step 1. Set A1 = A0 . Select aspect k1 = 1 with probability
q411 A1 5 =
1
0
1 + 2 + 3 + 4
Set 415 < 4j5, for j = 21 31 4, to indicate that this EBA
instance selects aspect 1 before aspects 2, 3 and 4. Eliminate alternative 3 because it does not have aspect 1. Set
C1 = C0 \839 = 811 29.
Step 2. Do not consider aspect 2 since it does not appear
in alternatives 1 and 2. Set A2 = 831 49. Select aspect k2 = 3
with probability
q431 A2 5 =
3
0
3 + 4
Set 435 < 445 because this EBA instance selects
aspect 3 before aspect 4.
Eliminate alternative 2 because it does not have aspect 3.
Set C2 = C1 \829 = 8190
Termination step. Stop because —C2 — = 1. Choose alternative 1.
The preceding EBA instance selects alternative 1 with
probability
3
1
·
0
q411 A1 5 · q431 A2 5 =
1 + 2 + 3 + 4 3 + 4
The precedence conditions are 415 < 425, 415 < 435,
415 < 445, and 435 < 445.
Next, consider the EBA instance in which Step 1 still
selects aspect k1 = 1 with probability q411 A1 5, but Step 2
selects aspect k2 = 4 from the set A2 = 831 49 with
probability
q441 A2 5 =
4
0
3 + 4
Then C2 = C1 \819 = 829. As C2 has a single alternative,
EBA stops after Step 2 and chooses alternative 2 with
probability
q411 A1 5 · q441 A2 5 =
4
1
·
0
1 + 2 + 3 + 4 3 + 4
Since aspect 1 is chosen from A1 = 811 21 31 49 and aspect 4
is chosen from A2 = 831 49, the precedence conditions are
415 < 425, 415 < 435, 415 < 445, and 445 <
435.
Finally, consider the three other instances, in which
aspect 2, 3, or 4 is selected in the first stage. In each
instance, EBA eliminates two of the three alternatives in
Step 1 and then stops. Specifically
(i) Step 1 selects aspect k1 = 2 with probability
q421 A1 5 =
2
0
1 + 2 + 3 + 4
It sets 425 < 4k5, for k = 11 31 4; eliminates alternatives 1 and 2; sets C1 = C0 \811 29 = 839; and chooses alternative 3 with probability q421 A1 5.
(ii) Step 1 selects aspect k1 = 3 with probability
q431 A1 5 =
3
0
1 + 2 + 3 + 4
It sets 435 < 4k5, for k = 11 21 4; eliminates alternatives 2 and 3; sets C1 = C0 \821 39 = 819; and chooses alternative 1 with probability q431 A1 5.
(iii) Step 1 selects aspect k1 = 4 with probability
q441 A1 5 =
4
0
1 + 2 + 3 + 4
It sets 445 < 4k5, for k = 11 21 3; eliminates alternatives 1 and 3; sets C1 = C0 \811 39 = 829; and chooses alternative 2 with probability q441 A1 5.
Across instances, EBA selects the alternatives x1 1 x2 1 and
x3 with the following probabilities:
p4x1 1 C5 = q411 A1 5 · q431 A2 5 + q431 A1 5
1
3
=
·
1 + 2 + 3 + 4 3 + 4
3
+
1 + 2 + 3 + 4
1
2
=
1−
1
1 + 4 /3
1 + 2 + 3 + 4
p4x2 1 C5 = q411 A1 5 · q441 A2 5 + q441 A1 5
1
4
=
·
1 + 2 + 3 + 4 3 + 4
4
+
1 + 2 + 3 + 4
1
2
=
1−
1
1 + 3 /4
1 + 2 + 3 + 4
2
p4x3 1 C5 = q421 A1 5 =
0
1 + 2 + 3 + 4
(2)
(3)
(4)
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516
Operations Research 63(3), pp. 512–526, © 2015 INFORMS
Observe that p4x1 1 C5 = a81 − p4x3 1 C59, p4x2 1 C5 =
41 − a581 − p4x3 1 C59, where a = 3 /43 + 4 5. That is, the
choice probability 1−p4x3 1 C5 of not selecting the Debussy
album is split between the two Beethoven albums in direct
proportion to the importance weights 3 and 4 . Suppose
the person is indifferent between the two Beethoven albums
and associates the same weight 3 = 4 = . Then a = 1/2
and
p4x1 1 C5 = p4x2 1 C5 =
p4x3 1 C5 =
1 − p4x3 1 C5
1
2
2
0
1 + 2 + 2
As  becomes smaller, the choice probabilities approach
the limiting values
1
2
lim p4x1 1 C5 = lim p4x2 1 C5 =
1−
1
→0
→0
2
1 + 2
2
0
lim p4x3 1 C5 =
→0
1 + 2
Thus, unlike the logit model, EBA considers the essential
choice to be between Beethoven and Debussy.
To see how EBA allows violations of order independence, suppose that 1 /2 = 1 + …, where … > 0 is a small,
positive value. Then
p4x3 1 C5 =
2
1
1
1
=
=
= − 2„1
1 + 2 1 + 1 /2 2 + … 2
where „ = …/48 + 4…5 > 0. Thus,
1
2
p4x1 1 C5 = p4x2 1 C5 =
1−
2
1 + 2
1 1
1
=
+ 2„ = + „0
2 2
4
2.2. Probabilistic Lexicographic Rule
On the other hand, if x2 were not available, then alternative x1 would be selected with probability 1 − p4x3 1 C5 =
1/2 + 2„. Thus, x1 has a higher choice probability than x3
when x2 is not available, but it has a lower choice probability than x3 when x2 is available. This is a violation of
order independence, which is captured by EBA but not by
the multinomial logit choice model.
Equation (1) gives the probability of selecting an aspect
at stage j of an EBA instance. The probability of an EBA
instance across its m stages is given by
m
Y
q4kj 1 Aj 5 =
j=1
j=1
kj
m
Y
P
k∈Aj
k
0
(5)
McFadden (1981) noted that since q4kj 1 Aj 5 has the form of
the Luce choice rule, the probability of selecting an aspect
at stage j can be written as
q4kj 1 Aj 5 = P
e
vkj
k∈Aj
e vk
1
where vk = ln4k 5 for all k ∈ A. Thus, if we restrict attention to a single stage of EBA, then the values of the aspect
choice probabilities can be trivially obtained by assuming that aspect k ∈ A has random utility uk = vk + …k ,
where vk is a deterministic component and …k is a stochastic component with extreme value distribution. The expression in Equation (5) follows immediately if we are willing
to assume that the utility of an aspect is an independent
draw from its distribution at each stage of EBA. That is,
if k1 is selected in the first stage with probability q4k1 1 A1 5,
then k2 will be selected in the second stage with probability q4k2 1 A2 — k1 1 A1 5 = q4k2 1 A2 5. However, as noted in the
introduction, Tversky’s view was that the aspect ordering
can change only across choice occasions, but that it is fixed
on a given choice occasion.
Since any of the n! possible orderings of the n aspects
is feasible, the derivation of EBA choice probabilities
requires considering each such ordering and showing that
an extreme value distribution of aspect utilities still yields
the EBA choice probabilities. We obtain this result in two
steps. The first step considers a probabilistic lexicographic
rule in which (i) each aspect has an independent, random
utility with an extreme value distribution, (ii) a single draw
from the utility distribution is obtained for each aspect on
a given choice occasion, and (iii) the aspects are ordered
in decreasing order of their utilities and then used in a
lexicographic rule to select an alternative from a choice
set. This yields an expression for the choice probabilities
associated with a subset of aspect sequences that eliminate
the same alternatives in the same order when used in a
lexicographic rule. The second step shows that each such
subset of sequences corresponds to an EBA instance and is
selected with a probability given by Equation (5).
A lexicographic rule uses a sequence of aspects to eliminate alternatives until a single alternative is selected. A
probabilistic lexicographic rule differs from a deterministic
lexicographic rule in only one way: it uses a probabilistic
mechanism to generate the sequence of aspects used in a
lexicographic rule. We first describe the probabilistic mechanism for generating sequences of aspect and then provide
a formal description of a lexicographic rule.
Let uk = vk + …k denote the utility a person associates
with aspect k, where vk is a deterministic component,
and …k is a stochastic component, for all k ∈ A = 811 0 0 0 1 n9.
The stochastic component represents uncertainty in a person’s evaluation of an aspect’s utility. It can occur because
the same aspect can appear more or less important, depending on the choice context or a person’s frame of mind.
For example, sometimes a person might have the taste for
meat, at other times the taste for a vegetarian meal. Tversky
and Sattah (1979) observed that such uncertainty can persist even when the effects of learning, satiation, or changes
in tastes are taken into account. They noted that even in
unique choice situations, where a choice is essentially made
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Operations Research 63(3), pp. 512–526, © 2015 INFORMS
only once, people can express doubt about the importance
they associate with one or another aspect of the alternatives.
For example, a person might not be sure how much he or
she values a larger screen size in a smart phone or television set. Such uncertainty cannot be reflected by a standard
error model, in which the total utility of an alternative is
allowed to be uncertain, but not the utilities of individual
aspects.
We consider a model in which …k , the stochastic component of aspect k’s utility, has an independent, extreme value
distribution, with density function fk 4…k 5 = exp4−…k e−…k 5
and cumulative distribution function Fk 4…k 5 = exp4−e−…k 50
On a given choice occasion, a single value of uk is obtained
as an independent random draw from its distribution, for
all k ∈ A. Let uk1 > · · · > ukn , where kj ∈ A for all j =
11 0 0 0 1 n. Let s = 4k1 1 0 0 0 1 kn 5 denote a sequence representing a preference ordering of the aspects for a person who
prefers aspect kj to aspect kj 0 , for all 1 ¶ j < j 0 ¶ n. We
say that sequence s has jth element kj , or equivalently, that
it assigns aspect kj to position j, for all j = 11 0 0 0 1 n.
Let S denote the set of n! distinct sequences. A lexicographic rule uses a sequence s ∈ S to choose an alternative
from C by first eliminating alternatives that do not have
aspect k1 , then eliminating those remaining alternatives that
do not have aspect k2 , and so on, until only one alternative
remains. An implementation of the rule terminates in at
most n steps and uses an aspect only if it appears in some,
but not all, of the remaining alternatives at a given step.
For brevity, we say that a sequence selects an alternative
from a choice set C, although strictly speaking it is a lexicographic rule that selects an alternative using the ordering
of aspects in a sequence. The following algorithm gives
a formal description of a lexicographic rule. The input to
the algorithm is a choice set C and a particular preference
ordering s = 4k1 1 0 0 0 1 kn 5 over the aspects.
Initialization step: Let s = 4k1 1 0 0 0 1 kn 5, C0 = C and xi =
4xi1 1 0 0 0 1 xin 5 for all i ∈ C.
Iteration step j 4¾ 15: Set Bj = 8i ∈ Cj−1 — xikj = 19,
(
Cj−1 if Bj = ™ or Bj = Cj−1 1
Cj =
Bj
otherwise,
and
(
0
tj =
1
if Bj = ™ or Bj = Cj−1 1
otherwise0
Termination step: Stop if —Cj — = 1. Set tj+1 = · · · = tn = 0
if the algorithm stops at the end of step j < n.
In the preceding algorithm, (i) Bj denotes the subset of
alternatives in Cj−1 that have aspect kj ; and (ii) tj = 1
indicates that aspect kj appears in some, but not all, of
the alternatives in Cj−1 ; otherwise, tj = 0. We specify the
values of tj because they will be useful for establishing
the equivalence between EBA and the probabilistic lexicographic rule. We illustrate the use of aspect sequences to
implement a lexicographic rule and the assignment of the
tj values with an example.
517
Example 2. We reconsider Example 1 in which C =
8x1 1 x2 1 x3 9, where x1 and x2 denote the two Beethoven
albums, and x3 denotes the Debussy album. Albums have the
following aspects: x1 has aspects 1 and 3. x2 has aspects 1
and 4, and x3 has aspect 2. Consider a lexicographic rule
using the aspect sequence s = 42 1 3 45. It uses aspects
k1 = 2, k2 = 1 and k3 = 3 one after another to eliminate
alternatives until only a single alternative remains. Since
there is only one Debussy album, this sequence stops after
one step. Thus, t1 = 1, t2 = t3 = t4 = 0, where
(i) t1 = 1 because aspect 2 (Debussy suite) is used
to eliminate both Beethoven albums from the set C0 =
8x1 1 x2 1 x3 9; and
(ii) t2 = t3 = t4 = 0 because the algorithm stops after
step 1 (C1 = 8x3 9).
Next, consider a lexicographic rule using the aspect
sequence s = 412345. It eliminates the Debussy album in
the first step and then eliminates the second Beethoven
album. Thus, t1 = 1, t2 = 0, t3 = 1, t4 = 0, where
(i) t1 = 1 because aspect 1 (Beethoven symphony)
appears in two of the three alternatives in C0 = 8x1 1 x2 1 x3 9;
(ii) t2 = 0 because aspect 2 (Debussy suite) appears
in neither of the alternatives in C1 = 8x1 1 x2 9 (the two
Beethoven recordings);
(iii) t3 = 1 because aspect 3 (which is unique to the first
Beethoven recording) appears in one of the two alternatives
in C1 = 8x1 1 x2 9; and
(iv) t4 = 0 because the algorithm stops after step 2
(C2 = 8x1 9).
Table 1 shows the lexicographic choices obtained by
each of the 4! = 24 aspect sequences. Note that an alternative can be chosen by more than one sequence.
To conclude, we note that although the aspect utilities
have independent distributions in the probabilistic lexicographic model, they are perfectly correlated across elimination stages. This is because the values of the aspect utilities
can only vary across choice occasions, but they are fixed
across elimination stages on a choice occasion. Similarly,
although the aspect utilities are independent, the choice
probabilities of the alternatives are not independent. The
reason is that if a (proper) subset of alternatives in a choice
set has common aspects, then the choice probabilities of
each alternative in the subset depends on the utilities of
these common aspects.
3. Equivalence of EBA and Probabilistic
Lexicographic Rule
We now establish the equivalence between the probabilistic
lexicographic rule and EBA. First we partition the set of all
possible aspect orderings into mutually exclusive and collectively exhaustive subsets. Table 1 shows this partitioning for the Beethoven-Debussy example. Aspect orderings
in each such subset are equivalent in the following sense:
when used by an individual in a lexicographic rule, they
each eliminate alternatives in any choice set in the same
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518
Operations Research 63(3), pp. 512–526, © 2015 INFORMS
Table 1.
Choices associated with lexicographic
sequences.
k1
k2
k3
k4
Choice
1
1
1
2
3
3
3
2
4
4
4
2
x1
1
1
1
2
4
4
4
2
3
3
3
2
x2
2
2
2
2
2
2
1
1
3
3
4
4
3
4
1
4
1
3
4
3
4
1
3
1
x3
3
3
3
3
3
3
1
1
2
2
4
4
2
4
1
4
1
2
4
2
4
1
2
1
x1
4
4
4
4
4
4
1
1
2
2
3
3
2
3
1
3
1
2
3
2
3
1
2
1
choice set. A proof for Lemma 1, and proofs for subsequent
lemmas and theorems, are provided in the appendix.
Lemma 1. Let sequence s = 4k1 1 0 0 0 1 kn 5 select alternative
i ∈ C using m steps, where 1 ¶ m < n. Let tj = 1 for j =
j1 1 0 0 0 1 jm and tj = 0 otherwise, where 0 < j1 · · · < jm . Then
sequence s 0 , which assigns position ‘4kj 5 to aspect kj , also
selects alternative i ∈ C using m steps, eliminating alternatives in the same order as sequence s, if ‘4kjr 5 < ‘4kj 5,
for all jr + 1 ¶ j ¶ n and 1 ¶ r ¶ m0
Lemma 1 is a key result for obtaining the correspondence
between EBA and a probabilistic lexicographic rule. We
illustrate the result with an example.
Example 3. Returning to Example 2, recall that the sequence s = 421345 chooses the Debussy album (x3 ) after
one step, so that t1 = 1, t2 = t3 = t4 = 0. Thus, all sequences
in which aspect 2 precedes all other aspects (that is, ‘425 <
‘4j5, for j = 11 31 4) also select the Debussy album. The
set of all such sequences is given by
K1 = 84213451 4214351 4231451 4234151 4241351 42431590
x2
Notes. x1 and x2 denote two different recordings of a Beethoven symphony; x3 denotes a recording of a Debussy suite. All aspect orderings in a
partition correspond to a set of sequences that satisfy the same precedence
conditions resulting in the choice of the same alternative.
sequence, using the same subset of aspects in the same
order. Then we show that there is a one-to-one mapping
between the probabilities associated with each subset of
equivalent aspect orderings and the EBA choice probabilities. The mapping establishes the desired equivalence and
implies that EBA can be characterized by random aspect
utilities with independent, extreme value distributions.
Recall that tj = 1405 if aspect kj is used (not used) by
the probabilistic lexicographic rule for eliminating alternatives at stage j. Let tj = 1 for m aspects, where 1 ¶ m ¶ n.
If m = n, then the choice set has at least n alternatives,
and sequence s = 4k1 1 0 0 0 1 kn 5 uses all n aspects to choose
alternative i ∈ C; another sequence s 0 = 4k10 1 0 0 0 1 kn0 5 may
also select i ∈ C, but only by eliminating alternatives in a
different order than sequence s. But if tj = 0 for some j =
11 0 0 0 1 n, then there can be other sequences s 0 that choose
alternative i ∈ C by eliminating alternatives in the same
order as sequence s. Suppose such a sequence s 0 exists.
Let aspect kj , which appears in position j in sequence s,
appear in position ‘4kj 5 in sequence s 0 , for all j = 11 0 0 0 1 n
(for example, suppose s = 43 1 25 and s 0 = 42 3 15; then
‘435 = 2, ‘415 = 3 and ‘425 = 1). Lemma 1 gives the condition under which two sequences, s and s 0 , use the same
aspects, in the same order, to eliminate alternatives from a
Similarly, recall that the sequence s = 412345 chooses
the first Beethoven album (x1 ) after two steps, and that
t1 = 1, t2 = 0, t3 = 1, t4 = 0. Thus, all sequences in which
aspect 1 precedes all other aspects (that is, ‘415 < ‘4j5,
for j = 21 31 4), and in which aspect 3 also precedes aspect
4 (that is, ‘435 < ‘445), select the same Beethoven album.
The set of all such sequences is given by
K2 = 84123451 4132451 41342590
In Table 1, the third grouping of sequences corresponds to
the set K1 , and the first grouping of sequences corresponds
to the set K2 . Other groupings in Table 1 correspond to
similar sets of sequences, each sequence in a set eliminating
the same alternatives, using the same aspects in the same
order.
As Example 3 illustrates, the precedence conditions
‘4kj 5 < ‘4k5 collectively specify the subset of sequences
K ⊂ S that eliminate the same alternatives, in the same
order, as sequence s ∈ S. We derive the probability of
obtaining the sequences in K and show that it is equal to
the probability of a corresponding EBA instance.
Consider a lexicographic rule that uses sequence s =
4k1 1 0 0 0 1 kn 5 and terminates in m ¶ n steps. By definition,
uk1 > · · · > ukn , and
K = 8s ∈ S — ‘4kjr 5 < ‘4kj 51 for all jr + 1 ¶ j ¶ n
and 1 ¶ r ¶ m91
where each sequence s ∈ K assigns position ‘4kj 5 to
aspect kj and satisfies the condition ‘4kjr 5 < ‘4kjr +1 5 <
· · · < ‘4kn 5, for each 1 ¶ r ¶ m. From Lemma 1, a lexicographic rule using any sequence s ∈ K eliminates alternatives in the same order, terminates in m steps, and
Kohli and Jedidi: EBA Error Theory
519
Operations Research 63(3), pp. 512–526, © 2015 INFORMS
selects the same alternative. The probability of obtaining a
sequence in K is given by
p4K5 =
X
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for all jr + 1 ¶ j ¶ n and r = 11 0 0 0 1 m51
where p4s5 is the probability of obtaining sequence s ∈ K.
Recall that uk = vk + …k denotes the utility a person associates with aspect k, where vk is a deterministic component,
and …k is a stochastic component with an extreme value distribution. Following Beggs et al. (1981), if uk1 > · · · > ukn ,
then sequence s = 4k1 1 0 0 0 1 kn 5 is obtained with probability
p4s5 =
v
n−1
Y
j=1
e
vk j
e kj
0
+ · · · + e vk n
(6)
The following example illustrates the computation of
p4K5 and shows how it relates to the EBA choice probabilities in Example 1.
Example 4. In Example 3, each sequence in the set
K1 = 84213451 4214351 4231451 4234151 4241351 42431591
selected the Debussy album (x3 ), and each sequence in the
set
K2 = 84123451 4132451 41342591
selected the first Beethoven album (x1 ). The probability of
obtaining a sequence in K1 is given by the sum of the
probabilities in Equation (6) for each sequence s ∈ K1 :
p4K1 5 =
e v2
1
e v2 + e v3 + e v4 + e v5
e v3
e v1
0
p4K2 5 = v
v
v
v
v
e 1 + e 2 + e 3 + e 4 e 3 + e v4
p4K1 5 =
p4s5 = p4‘4kjr 5 < ‘4kj 51
s∈K
Simplifying the above expressions gives
e v1
e v3
e v2
·
·
ev1 + ev2 + ev3 + ev4 ev1 + ev3 + ev4 ev3 + ev4
e v1
e v2
e v4
·
+ v
·
e 1 + e v2 + e v3 + e v4 e v1 + e v3 + e v4 e v3 + e v4
e v2
e v3
e v1
+ v
· v
· v
v
v
v
v
v
e 1 + e 2 + e 3 + e 4 e 1 + e 3 + e 4 e 1 + e v4
e v2
e v3
e v4
· v
· v
+ v
v
v
v
v
v
e 1 + e 2 + e 3 + e 4 e 1 + e 3 + e 4 e 1 + e v4
e v2
e v4
e v1
+ v
·
·
e 1 + e v2 + e v3 + e v4 e v1 + e v3 + e v4 e v1 + e v3
e v2
e v4
e v3
+ v
·
0
·
e 1 + e v2 + e v3 + e v4 e v1 + e v3 + e v4 e v1 + e v3
Similarly, the probability of obtaining a sequence in K2 is
given by
e v2
e v3
e v1
p4K2 5 = v
· v
· v
v
v
v
v
v
e 1 + e 2 + e 3 + e 4 e 1 + e 3 + e 4 e 3 + e v4
e v1
e v3
e v2
+ v
· v
· v
v
v
v
v
v
e 1 + e 2 + e 3 + e 4 e 1 + e 3 + e 4 e 2 + e v4
e v1
e v3
e v4
·
·
0
+ v
e 1 + e v2 + e v3 + e v4 e v3 + e v3 + e v4 e v2 + e v4
Thus, (i) p4K1 5 is the multinomial logit probability of
choosing aspect 2 from aspects 1–4, and (ii) p4K2 5 is the
product of two multinomial logit probabilities, one corresponding to the choice of aspect 1 from aspects 1–4, and
the other corresponding to the choice of aspect 3 from
aspects 3 and 4. If we set evk = k in the above expressions, we obtain a value of p4K1 5 that is identical to the
probability p4x3 1 C5 in Equation (4) for an EBA instance
that terminates after a single step upon choosing aspect 2.
Similarly, we obtain a value of p4K2 5 that is identical to the
probability p4x2 1 C5 in Equation (3) for an EBA instance
which (i) uses aspect 1 in step 1, (ii) ignores aspect 2, and
(iii) uses aspect 3 in step 2, after which it terminates.
The equivalence between the probabilities of EBA instances and sets of lexicographic sequences in Example 4
is valid in general. We show the equivalence without the
cumbersome enumeration of equivalent sequences characterizing a set K. We do so by integrating over the distributions of aspect utilities in such a way that all precedence
constraints of the form ‘4kjr 5 < ‘4kj 5 that characterize a
set K are satisfied.
Since uk ¾ uk0 when ‘4k5 < ‘4k0 5, we can write p4K5 as
p4K5 = p4‘4kjr 5 < ‘4kj 51 for all jr + 1 ¶ j ¶ n
and r = 11 0 0 0 1 m5
= p4ukjr ¾ ukj 1 for all jr + 1 ¶ j ¶ n
and r = 11 0 0 0 1 m50
(7)
For example, if m = 2, then Equation (7) can be written as
p4K5 = p4ukj ¾ ukj 1 for all j1 + 1 ¶ j ¶ n and ukj ¾ ukj 1
1
2
for all j2 + 1 ¶ j ¶ n50 (8)
Since ukj ¾ ukj , the condition ukj ¾ ukj implies ukj ¾ ukj
1
2
2
1
for all j2 + 1 ¶ j ¶ n. Thus, we can write Equation (8) as
p4K5 = p4ukj ¾ ukj 1 for all j1 + 1 ¶ j ¶ j2 and ukj ¾ ukj 1
1
2
for all j2 + 1 ¶ j ¶ n50 (9)
More generally, ukj ¾ ukj ¾ · · · ¾ ukjm implies that
1
2
Equation (7) can be written as
p4K5 = p4ukjr ¾ ukj 1 for all jr + 1 ¶ j ¶ jr+1
and r = 11 0 0 0 1 m51
where, by definition, jm+1 = n.
(10)
Kohli and Jedidi: EBA Error Theory
520
Operations Research 63(3), pp. 512–526, © 2015 INFORMS
Let ukj = ˆ. Then for any r = 11 0 0 0 1 m,
Lemma 2.
0
p4ukjr ¾ ukj 1 for all jr + 1 ¶ j ¶ jr+1 5
Z ukj Z ukj
Z ukj
r
r
r−1
=
000
fkjr 4ukjr 5fkjr +1 4ukjr +1 5
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−ˆ
−ˆ
· · · fkj
r+1
Ir =
5 dukjr dukjr +1 · · · dukj
r+1
r+1
r−1
fkjr 4ukjr 5
−ˆ
Z
Y
j=jr +1
Fkj 4ukjr 5 dukjr 1
(11)
u
fkj 4ukj 5 dukj
−ˆ
is the cumulative distribution function for ukj = vkj + …kj ,
the utility of aspect kj . Let
jr+1 −1
Gr = fkjr 4ukjr 5
Y
j=jr +1
F kj 4ukjr 51
for all r = 11 0 0 0 1 m − 11
and
jm+1
Gm = fkjm 4ukjm 5
Y
j=jm +1
F kj 4ukjr 51
where jm+1 = n. Then Equation (11) can be written as
p4ukjr ¾ ukj 1 for all jr + 1 ¶ j ¶ jr+1 5
Z ukj
Z ukj
r
r−1
fkj 4ukj 5 dukj
Gr
=
r+1
r+1
−ˆ
−ˆ
r+1
dukjr 1
(12)
for any r = 11 0 0 0 1 m. Thus,
p4K5 = p4ukjr ¾ ukj 1 for all jr +1 ¶ j ¶ jr+1
and r = 110001m5
0
1
ukj
1
···
Z
m−1
0
ukj =−ˆ
ukj
2
Y
Fkj 4ukj 5
2
j=j2 +1
jm+1
Y
f 4ukjm 5
Fkj 4ukjm 5 dukjm ···dukj dukj
G1
1
Z
1
2
j=jm +1
ukj
···
Fkj 4ukj 5
j=j1 +1
fkj 4ukj 5
2
ukj
j3 −1
−ˆ
=
1
−ˆ
Z
Y
fkj 4ukj 5
−ˆ
Z
j2 −1
ukj
Z
·
vkj
j=jr+1
r
e
vkj
n
X
−ukj
vkj
r−1
exp −e
1
e
j=jr
for all r = 11 0 0 0 1 m − 10
Theorem 1.
p4K5 =
where
=
n
X
−u
v
Gr exp −e kjr
e kj dukjr
m
Y
wkjr
r=1
wkjr + · · · + wkjn
1
(14)
jr+1
ukj
Fkj 4u5 =
ukj =−ˆ
j=jr
p4ukjr ¾ ukj 1 for all jr + 1 ¶ j ¶ jr+1 5
Z
r−1
e
= Pn
1
where the first term integrates ukjr over 4−ˆ1 ukj 5, and
r−1
each subsequent term integrates ukj over 4−ˆ1 ukjr 5, for
each j = jr + 11 0 0 0 1 jr+1 . Since the aspect utilities are independent, we can rewrite the above expression as
=
ukj
r
−ˆ
4ukj
Z
m−1
ukj =−ˆ
Z
ukj
1
ukj =−ˆ
1
G2
2
Gm dukjm ···dukj dukj 0
2
1
(13)
m
The following lemma is useful for proving Theorem 1,
which gives a closed-form expression for p4K5.
where wk = evk , for all k = 11 0 0 0 1 n.
Observe that if m = n, then K contains only one
sequence, s = 4k1 1 0 0 0 1 kn 5. In this case, the expression in
Theorem 1 becomes identical to the probability in Equation (6), which is obtained by Beggs et al. (1981) for the
rank-ordered logit model.
Theorem 2. Elimination by aspects is equivalent to a
probabilistic lexicographic rule.
Theorem 2 implies that the weights in an EBA model
can be specified using a random utility model for aspects.
Aspect j has weight wj if it has random utility uj =
ln4wj 5 + …j , where …j is an independent observation with
extreme value distribution, for all j = 11 0 0 0 1 n. Note that
the logit model assumes independence from irrelevant alternatives. In the present instance, the rank-ordered logit
model selects aspects (not alternatives). In other words,
aspect choice is independent of irrelevant aspects. This
property allows EBA to violate order independence when
selecting an alternative from a choice set.
A probabilistic lexicographic rule describes a hierarchical model with n levels and n! branches. A path from
the root to a terminal node characterizes a lexicographic
sequence of aspects. Each nest is probabilistically constructed. For example, aspect j1 is selected at Pthe first
level of the nesting with probability pj1 = wj1 / nr=1 wjr ,
for all j1 = 11 0 0 0 1 n; then aspect j2 is selected at
P the second
level of nesting with probability pj2 — j1 = wj2 / nr=11 r6=j1 wr ,
for all j2 6= j1 , j2 = 11 0 0 0 1 n0 The nesting sequence concludes when each aspect has been considered. Elimination
by aspects prunes this tree by eliminating those aspects that,
given the preceding levels of the tree, are no longer relevant for choosing an alternative from a choice set. However,
this is equivalent to adding the occurrence probabilities for
those sequences that lead to the same choice, by the same
sequence of eliminations by aspect, in the probabilistic lexicographic rule.
Example 5. We reconsider Example 1 to illustrate the
equivalence of EBA and the probabilistic lexicographic
rule. Recall that we used four aspects to characterize each
of three alternatives: x1 = 41 0 1 05, x2 = 41 0 0 15, x3 =
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521
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Operations Research 63(3), pp. 512–526, © 2015 INFORMS
40 1 0 05, where x1 and x2 are the two recordings of a
Beethoven symphony, and x3 is a recording of a Debussy
suite.
Table 1 shows the alternative chosen by each sequence
using the probabilistic lexicographic rule. We explain these
choices below for alternative x1 .
(i) Alternative x1 is chosen by the set of sequences satisfying the conditions ‘415 < ‘4j5, for j = 21 31 4; and
‘435 < ‘445—that is, the sequences in which aspect 1 is
selected before any other aspect, and aspect 3 is selected
before aspect 4. This set of sequences is
K1 = 84123451 4132451 41342590
From Theorem 1, the probability of choosing x1 by using
a sequence in K1 is
w1
w3
p4K1 5 =
·
1
w1 + w2 + w3 + w4 w3 + w4
where wk = evk and vk is the deterministic component of the
utility for aspect k. Note that p4K1 5 is also the probability
of choosing x1 using an instance of EBA in which aspect 1
is first selected (with probability w1 /4w1 + w2 + w3 + w4 5),
aspect 2 is ignored (because it appears in neither of the
two Beethoven albums), and then aspect 3 is selected (with
probability w3 /4w3 + w4 5).
(ii) Alternative x1 is also chosen by the set of sequences
satisfying the conditions ‘435 < ‘4j5, for j = 11 21 4. This
set is
K2 = 84312451 4314251 4321451 4324151 4341251 43421590
From Theorem 1, the probability of choosing x1 by using
a sequence in K2 is
w3
p4K2 5 =
0
w1 + w2 + w3 + w4
Thus, the probabilistic lexicographic rule selects alternative x1 with probability
w1
w3
·
p4x1 5 = p4K1 5 + p4K2 5 =
w1 + w2 + w3 + w4 w3 + w4
w3
+
w1 + w2 + w3 + w4
w3
w1
=
1+
w1 + w2 + w3 + w4
w3 + w4
w2
w3
1−
0
=
w3 + w4
w1 + w2 + w3 + w4
Similarly, the probabilistic lexicographic rule selects alternative x2 with probability
w1
w4
p4x2 5 =
·
w1 + w2 + w3 + w4 w3 + w4
w4
+
w1 + w2 + w3 + w4
w1
w4
=
1+
w1 + w2 + w3 + w4
w3 + w4
w4
w2
=
1−
0
w3 + w4
w1 + w2 + w3 + w4
Finally, all six sequences satisfying the conditions ‘425 <
‘4j5, for j = 11 31 4, select alternative x3 . Thus, the choice
probability for alternative x3 is given by
p4x3 5 = 1 − p4x1 5 − p4x2 5 =
w2
0
w1 + w2 + w3 + w4
Let wj = j = ln4vj 5. Then p4xj 5 = p4xj 1 C5, where
p4xj 1 C5 is the EBA probability of choice for alternative j
in Example 1, given by Equations (2)–(4).
4. Probabilistic Utility Function for EBA
As noted in the introduction, Tversky (1972b) showed that
EBA is a random utility model, but did not obtain a utility
function for the alternatives or a statistical distribution over
the utilities. The equivalence between EBA and probabilistic lexicographic rules allows us to do so.
On a choice occasion, a person using a probabilistic lexicographic rule selects some sequence s ∈ S with probability p4s5. Any such sequence specifies a lexicographic preference ordering over the set of alternatives that are available
to the person. In turn, each lexicographic preference ordering can be represented by a utility function over the aspects.
We provide the details below.
Let C denote the set of all alternatives. Let A denote the
set of aspects associated with the alternatives in C. Let S
denote the set of all possible aspect orderings of these —A—
aspects.
Let C ⊆ C denote a subset of the —C— alternatives, and let
A ⊆ A denote the set of aspects associated with the alternatives in C, where —A— = n. As in the preceding sections, let
S denote the set of n! orderings of the aspects in A. Observe
that unless A = A, S is not a subset of the sequences in S
because each s ∈ S is defined over n aspects, whereas each
sequence in S is defined over —A— aspects.
Each sequence s = 4k1 1 0 0 0 1 kn 5 ∈ S specifies a lexicographic preference ordering over the alternatives in C ⊆ C.
Given any sequence s ∈ S, we wish to specify (i) a function that assigns a utility to each alternative in C, and (ii) a
probability with which the alternatives in C simultaneously
obtain these utility values.
Equation (6) gives the probability p4s5 that the probabilistic lexicographic rule uses the sequence s ∈ S. Kohli
and Jedidi (2007) showed that the ordering of the alternatives in C, obtained by a lexicographic rule using sequence
s = 4k1 1 0 0 0 1 kn 5, can be represented by the utility function
u4xi 5 =
xik1
21
+
xik2
22
+···+
xikn
2n
1
where xi = 4xi1 1 0 0 0 1 xin 5 is the profile of alternative i ∈ C,
and xik = 1405 if alternative i has (does not have) aspect k.
Note that each sequence s ∈ S simultaneously specifies the
utilities for all alternatives i ∈ C.
Since EBA does not use aspects that are common to all
alternatives in a choice set, any aspect k ∈ A\A appears
in either all or none of the alternatives in a choice set C.
Kohli and Jedidi: EBA Error Theory
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522
Operations Research 63(3), pp. 512–526, © 2015 INFORMS
However, it makes no difference to the ordering of the alternatives in C if we compute their utility values by using the
subset of aspects in A or by using all aspects in A. The
reason is that if aspect k appears in none of the alternatives in C (that is, if xik = 0 for all i ∈ C), then including
aspect k in the utility function does not change the utility
of any alternative i ∈ C. And if aspect k appears in all of
the alternatives in C (that is, if xik = 1 for all i ∈ C), then
including aspect k in the utility function changes the utility
of each alternative i ∈ C by the same value.
We now discuss the link between the utilities of alternatives and their EBA choice probabilities. Given a set
C ⊆ C, let p4xi 1 C5 denote the probability that a person
selects alternative i ∈ C on a choice occasion. Then
h
i
p4xi 1 C5 = Pr u4xi 5 = max u4xl 5 0
l∈C
Let
n
o
K4i5 = s ∈ S — i = arg max u4xl 5
l∈C
denote the subset of sequences in S for which alternative
i ∈ C has the highest utility, u4xi 5 = maxl∈C u4xl 5. Since an
alternative with a higher utility has a higher lexicographic
ranking, alternative i ∈ C is ranked first by each sequence
s ∈ K4i5 when the sequence is used by a lexicographic rule
to arrange the alternatives in C in decreasing preference
order. Let
K4i5 = K1 4i5 ∪ K2 4i5 ∪ · · · ∪ Kh 4i51
h ¾ 11
where each Kt 4i5 ⊆ K4i5, t = 11 0 0 0 1 h, is a subset of
sequences that, when used in a lexicographic rule, satisfy
the same precedence conditions; that is, ‘4kjr 5 < ‘4kj 5, for
all jr + 1 < j ¶ n and r = 11 0 0 0 1 m. Then, by definition, the
choice probability for alternative i ∈ C is
p4xi 1 C5 =
h
X
p4Kt 4i551
t=1
where p4Kt 4i55 is given by Theorem 1, and the summation
runs over the h subsets of sequences that choose the same
alternative i ∈ C.
Example 6. We reconsider Example 1. Let A = 811 21 31 49
denote a set of four aspects. Let S = 8411 21 31 451
0 0 0 1 441 31 21 159 denote the set of 4! = 24 sequences, which
are listed in Table 1. Let C = 811 0 0 0 1 159 denote the set
of 24 − 1 = 15 alternatives, each of which has one or more
aspects. Then A = A, S = S, and C = 811 21 39 ⊂ C, where
x1 = 41 0 1 05 and x2 = 41 0 0 15 are the two Beethoven
albums, and x3 = 40 1 0 05 is the Debussy album. Consider
the sequence s = 4k1 1 k2 1 k3 5, which is used in a probabilistic lexicographic rule with the probability
p4s5 =
k1
·
k2
·
k3
k1 + k2 + k3 + k4 k2 + k3 + k4 k3 + k4
0
Let
xik1
u4xi 5 =
2
+
xik2
22
+
xik3
23
+
xik4
24
denote the utility of alternative i ∈ C. Then the vector u =
4u4x1 51 u4x2 51 u4x3 55 is obtained with probability p4s5, for
all s ∈ S. For example, if s = 411 21 31 45, then
1
1
0
10
0
+ 2+ 3+ 4= 1
2 2
2
2
16
1
0
0
1
9
u4x2 5 = + 2 + 3 + 4 = 1
2 2
2
2
16
0
1
0
0
4
u4x3 5 = + 2 + 3 + 4 = 0
2 2
2
2
16
u4x1 5 =
We note that u4x1 5 > u4x2 5 > u4x3 5, which is consistent
with the preference ordering 11 21 3 associated with the
alternatives by a lexicographic rule using the aspect ordering s = 411 21 31 45. We say that the alternatives in C have
the utilities 4u4x1 51 u4x2 51 u4x3 55 = 410/161 9/161 4/165
with probability
p4s5 =
2
3
1
·
·
1
1 + 2 + 3 + 4 2 + 3 + 4 3 + 4
for s = 411 21 31 450
Similarly, if s = 431 21 41 15, then
1
0
1
9
0
+ + + = 1
2 22 23 24 16
0
1
1
3
0
u4x2 5 = + 2 + 3 + 4 = 1
2 2
2
2
16
1
0
0
4
0
u4x3 5 = + 2 + 3 + 4 = 0
2 2
2
2
16
u4x1 5 =
We note that u4x1 5 > u4x3 5 > u4x2 5, which is consistent with the preference ordering 11 31 2 of the alternatives obtained by a lexicographic rule using the sequence
s = 431 21 41 15. We say that the alternatives in C have the
utilities 4u4x1 51 u4x2 51 u4x3 55 = 49/161 3/161 4/165 with
probability
p4s5 =
3
2
4
·
·
1
1 + 2 + 3 + 4 1 + 2 + 4 1 + 4
for s = 431 21 41 150
Similar probabilities can be associated with the utility vectors for alternatives in any choice set C ⊆ C.
Table 1 shows the partitions (subsets of sequences) that
choose the same alternative from the choice set C =
811 21 39. For example, alternative 1 is selected by the first
and the fourth partitions shown in the table. Example 5
gives the choice probabilities for each alternative.
Kohli and Jedidi: EBA Error Theory
523
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5. Implications for Model Estimation
The error theory we developed in this paper implies that
EBA is a member of the logit family of models. Suppose
we choose to represent each alternative by a single, unique
aspect. Then each EBA instance stops after m = 1 step. The
EBA model becomes equivalent to the multinomial logit
model. Aspect utilities become indistinguishable from the
utilities of alternatives, and the probability of choosing an
aspect is equal to the probability of choosing an alternative.
Now suppose an EBA instance eliminates a single alternative at each step. Then it terminates in m = n steps, and its
occurrence probability is given by the rank-ordered logit
model. If 1 < m < n, the probability associated with an
EBA instance is a product of m terms, the jth term corresponding to the multinomial logit probability of choosing
aspect kj ∈ Aj at step j.
The choice probability for an alternative is obtained by
adding the probabilities of EBA instances that result in its
choice (see Equations (2)–(4) in Example 1). Since the
probability of an EBA instance is a function of the aspect
choice probabilities, the latter can be used to formulate a
likelihood function, which can be maximized to estimate
the parameters of an EBA model. As errors are associated
with aspects, heterogeneity and covariates can be introduced at the aspect level.
Heterogeneity in EBA rules can be specified by allowing
the deterministic aspect utilities, vk , to vary across individuals or segments. For example, a random effects model can
be formulated in which each individual i is considered to
associate utility vik = v̄k + ƒ 0 zi + ei with aspect k, where
v̄k is an intercept term, zi is a vector of individual-level
covariates (e.g., demographic variables), ƒ 0 is the associated vector of parameters, and ei has a Normal distribution
across individuals, with mean zero and variance ’k2 . This
formulation allows both observed and unobserved heterogeneity in the aspect utilities. A latent class formulation
can also be obtained by allowing each vik value to be a
2
draw from a mixture of Normal distributions N 4vgk 1 ’gk
5,
where g = 11 0 0 0 1 G is an index denoting segments.
Since aspects have utilities, their deterministic components can be further decomposed in the same way as the
utilities of alternatives are decomposed in a logit model.
For example, such a decomposition may be useful for an
aspect like the affordability of a durable good, which may
be a composite of its price, maintenance cost, and expected
life; or for an aspect like the convenience of a travel mode,
which may be a composite of ease of access, waiting time
and travel time. This reparametrization allows the modeling of aspect utilities as functions of other variables that
are common to subsets of aspects. Consider an aspect with
utility uk = v̄k +…k , where …k has the extreme value distribution. Suppose that aspect k is composed of other features;
that is,
v̄k = ‚k0 + ‚1 yk1 + · · · + ‚t ykt 1
where ykj are the values of the variables y1 1 0 0 0 1 yt associated with aspect k. The standard EBA model corresponds
to v̄k = ‚k0 for all k ∈ A.
6. Conclusion
Elimination by aspects is a model of bounded rationality
that is considered to represent consumer choice more accurately than models assuming utility maximizing consumers.
One limitation of the model has been the lack of an error
theory. We obtain an error theory and show that EBA can be
derived by assuming that aspects have random utilities with
independent, extreme value distributions. The result implies
that EBA is consistent with the assumption that consumers
maximize random utility when ordering the aspects they
use to eliminate alternatives from a choice set. EBA generalizes the multinomial logit and rank-ordered logit models.
Maximum likelihood methods currently available for logit
models can be modified to estimate EBA parameters. The
model can be extended to allow aspect utilities to be functions of covariates and to capture consumer heterogeneity
using latent-class or random-effects approaches.
The proposed error structure can also be associated with
aspects in a special case of EBA called preference trees
(Tversky and Sattah 1979). Since the latter have the same
hierarchical structure as nested logit models, it can be useful to compare the empirical performance of the two models. Conceptually, the proposed error structure implies that
preference trees do not require the assumption of a generalized extreme value distribution, which is needed for nested
logit models. Batley and Daly (2006) showed that there is
a mapping between the parameters of an EBA model and
a nested logit model when there are three alternatives and
two aspects. The problem they considered is equivalent to
the problem of choosing between Beethoven and Debussy
record albums discussed in the present paper. It may be
useful to examine if the proposed error theory can be useful for establishing a more general correspondence between
nested logit models and preference trees.
EBA uses the Luce choice rule for selecting aspects
at each stage of elimination. Since the Luce choice rule
assumes IIA, EBA assumes that aspect selection is independent of irrelevant aspects. It may be useful to examine
how the EBA model changes if the Luce choice rule is
replaced by another model for aspect selection. This can be
achieved by considering alternative statistical distributions
for aspect utilities. These distributions can be used to derive
the choice probabilities of alternatives using a probabilistic
lexicographic rule, and thus for an equivalent EBA model.
Acknowledgments
The authors are grateful to Omar Besbes and Khaled Boughanmi
for their help in proving Theorem 1.
Kohli and Jedidi: EBA Error Theory
524
Operations Research 63(3), pp. 512–526, © 2015 INFORMS
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Appendix. Proofs of Lemmas and Theorems
Lemma 1. Let sequence s = 4k1 1 0 0 0 1 kn 5 select alternative i ∈ C
using m steps, where 1 ¶ m < n. Let tj = 1 for j = j1 1 0 0 0 1 jm
and tj = 0 otherwise, where 0 < j1 · · · < jm . Then sequence s 0 ,
which assigns position ‘4kj 5 to aspect kj , also selects alternative i ∈ C using m steps, eliminating alternatives in the same
order as sequence s, if ‘4kjr 5 < ‘4kj 5, for all jr + 1 ¶ j ¶ n and
1 ¶ r ¶ m.
Proof. Consider sequence s. Let tj1 = · · · = tjm = 1, where 1 ¶
j1 < · · · < jm < n; and tj = 0, for all jr < j < jr+1 and r =
11 0 0 0 1 m (define jm+1 = n + 1). Then Cj = Cjr , for all jr < j <
jr+1 , because tj = 0 means that aspect kj is not used to eliminate alternatives. Since Cj ⊆ Cj−1 for all j ¾ 1, aspect kj cannot
appear in alternatives in the subsets Cjr+1 1 0 0 0 1 Cjm that are associated with sequence s.
Now consider sequence s 0 . If j1 > 1, then aspects k1 1 0 0 0 1 kj1 −1
appear in none or all of the alternatives in C and therefore are
not used to eliminate alternatives. Thus, these aspects can appear
at any position in s 0 without affecting the order in which s 0 eliminates alternatives from C. Similarly, if j2 > j1 + 1, then aspects
kj1 +1 1 0 0 0 1 kj2 −1 appear in none or all of the alternatives in C1 and
therefore are not used to eliminate alternatives. Thus, these aspects
can appear at any position in s 0 after aspect kj1 without affecting
the order in which s 0 eliminates alternatives from C. More generally, since ‘4kj1 5 < · · · < ‘4kjm 5, assigning position ‘4kj 5 >
‘4kjr 5 to aspect kj when jr < j < jr+1 , for all r = 11 0 0 0 1 m, has
no effect on the sequence in which s 0 eliminates alternatives from
C. It follows that sequence s 0 also eliminates alternatives from C
in the same order as sequence s. ƒ
Ir =
Z
ukj
r−1
j=jr+1
r
e
n
P
=
vkj
e
= Pn
vkj
r
vkj
e
j=jr
vkj
j=jr
f 4ukj 5
r
ukj =−ˆ
Y
Theorem 1.
p4K5 =
m
Y
wkj
r=1
wkj + · · · + wkj
r
r
Proof. We can write the expression for p4K5 as
p4K5 = p4ukj ¾ ukj 1 for all jr + 1 ¶ j ¶ n
r
and r = 11 0 0 0 1 m5
= p4ukj ¾ ukj 1 for all jr + 1 ¶ j ¶ jr+1
r
and r = 11 0 0 0 1 m5
Z u
Z u
Z ukj
kj
kj
2
1
0
G3
G2
G1
=
ukj
m−2
=−ˆ
m−1
Gm dukj
m
ukj =−ˆ
dukj
m−1
m
1
m
·
e
vkj
ukj
Z
0
ukj =−ˆ
G1
Z
Z
m−2
m−1
=−ˆ
−ukj
r
e
vkj
r
5e
−ukj
r
e
r
G2
Z
ukj
2
ukj =−ˆ
G3
3
n
X
−u
v
Gm−1 exp −e kjm−1
e kj dukj
m−1
j=jm
· · · dukj
vkj
1
2
ukj
uk j
uk j
ukj =−ˆ
1
Substituting the expressions for fk 4 · 5 and Fk 4 · 5 gives
exp4−e
ukj
vkj
j=jm
···
r−1
Gm−1
Z
First substitute4
v
Z ukj
n
X
e kjm
−ukj
vkj
m−1
m−1
Gm dukj = Pn
exp
−e
e
vkj
m
ukj =−ˆ
j=jm e
j=jm
m
r
j=jr+1
ukj
3
2
1
Z
ukj =−ˆ
ukj =−ˆ
ukj =−ˆ
r
n
X
−u
v
· exp −e kjr
e kj dukj 0
Z
1
n
where wk = evk , for all k = 11 0 0 0 1 n.
e
p4K5 = Pn
F kj 4ukj 5
j=jr +1
r
Ir =
j=jr
in the preceding expression for p4K5 to obtain
jr+1 −1
r−1
−ˆ
j=jr
vkj
e
uk =uk
n
jr−1
jr
X
−u
v
exp −e kjr
e kj
n
X
−u
v
exp −e kjr−1
e kj 0 ƒ
r
j=jr
vkj
dukj
2
Proof. By definition,
Ir =
e
= Pn
e
r
r
for all r = 11 0 0 0 1 m − 10
ukj
vkj
j=jr
· · · dukj dukj
j=jr
Z
e
j=jr
m−1
n
X
−u
v
e kj 1
exp −e kjr−1
r
e
n
X
ukj
n
X
−u
v
Gr exp −e kjr
e kj dukj
ukj =−ˆ
vkj
·
···
Lemma 2.
Pn
Assembling terms and multiplying and dividing by
gives
v
Z uk j
n
e kjr
−ukj X vkj
−u
r−1
r
exp
−e
e
e kjr
Ir = Pn
vkj
e
=−ˆ
u
k
j=jr
j=jr
jr
2
dukj 0
1
ukj =−ˆ
r
jr+1 −1
n
X vk
X
−u
−u
v
×exp −e kjr
e j exp −e kjr
e kj dukj
j=jr+1
j=jr +1
=e
vkj
Z
uk j
r−1
r
exp4−e
−ukj
r
e
vkj
r
5e
r
−ukj
r
ukj =−ˆ
Next, use Lemma 2 to make a sequence of m − 1 substitutions in
the expression for p4K5. The substitutions begin with r = m − 1
and end with r = 1. Thus,
Z uk j
n
X
−u
v
m−2
Im−1 =
Gm−1 exp −e kjm−1
e kj dukj
ukj
r
=−ˆ
m−1
jr+1 −1
n
X vk
X
−u
−u
v
e kj dukj 0
×exp −e kjr
e j exp −e kjr
j=jr +1
j=jr+1
r
e
= Pn
vkj
m−1
j=jm−1
e
vkj
j=m
n
X
−u
v
· exp −e kjm−2
e kj 0
j=jm−1
m−1
Kohli and Jedidi: EBA Error Theory
525
Operations Research 63(3), pp. 512–526, © 2015 INFORMS
Making this substitution in the preceding expression for p4K5
gives
e
vkj
p4K5 = Pn
e
j=jm−1
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·
e
m−1
vkj
ukj
Z
0
G1
ukj =−ˆ
···
j=jm
Z
e
vkj
uk j
1
ukj =−ˆ
G2
Z
m−3
m−2
=−ˆ
ukj
2
ukj =−ˆ
G3
3
2
ukj
ukj
m
· Pn
1
Z
vkj
Gm−2
n
X
−u
v
· exp −e kjm−2
e kj dukj
m−2
j=jm−1
0 0 0 dukj dukj 0
2
1
Recursive substitutions using Lemma 2 for r = m − 21 0 0 0 1 1, in
the expression for p4K5 yield
p4K5 =
m
Y
r=1
e
Pn
vkj
e
j=jr
=
vkj
wkj
m
Y
r
r
r=1 wkj + · · · + wkj
r
1
where wk = e , for all k = 11 0 0 0 1 n.
ƒ
Theorem 2. Elimination by aspects is equivalent to a probabilistic lexicographic rule.
Proof. Consider an instance of EBA that selects aspect kj ∈ Aj .
By definition,
4kj 5 < 4k51
for all k ∈ Aj \8kj 9 and j = 11 0 0 0 1 m3
and
q=
m
Y
q4kj 1 Aj 5 =
j=1
kj
m
Y
j=1
P
k∈Aj
k
0
Let K denote the set of sequences that satisfy
‘4kj 5 < ‘4k51
if 4kj 5 < 4k51 for all k ∈ Ak \8kj 90
From Lemma 1, each sequence s ∈ K eliminates alternatives in
the same order and chooses the same alternative i ∈ C, in m steps,
where 1 ¶ m ¶ n. Theorem 1 implies that
p4K5 =
wkj
m
Y
j=1
P
wk
k∈Aj
0
It follows that q = p4K5 if k = wk , for all k ∈ A.
Conversely, suppose K ⊆ S is a subset of sequences that satisfies
‘4kjr 5 < ‘4kj 5
for all jr + 1 ¶ j ¶ n and r = 11 0 0 0 m0
From Theorem 1,
m
wkj
Y
p4K5 =
0
w
+
· · · + wkn
j=1 kr
Consider
Ar = 8kjr 1 kjr +1 1 0 0 0 1 kn 91
for all r = 11 0 0 0 1 m0
Then each sequence s ∈ K eliminates alternatives in the same
order, and select the same alternative i ∈ C, as an instance of EBA
that selects aspect kjr ∈ Ar at step r, for all r = 11 0 0 0 1 m. Thus,
there is an instance of EBA that selects alternative i ∈ A with
probability
q=
kj
m
Y
j=1
P
k∈Aj
k
0
It follows that p4K5 = q if wk = k , for all k ∈ A.
1. Order independence is equivalent to simple scalability, which
requires the choice probabilities of alternatives to be monotone
functions of their scale values.
2. Note that a fixed aspect ordering does not mean that the probability of selecting an aspect is constant across stages. Instead,
it depends upon the relative importances of the aspects that are
available at any given stage of EBA.
3. The same example is sometimes described as the red-bus/bluebus problem, in which a person chooses between a car (Debussy
suite) and two equally good buses (Beethoven symphony), one
with red color (album 1) and the other with blue color (album 2).
4. The evaluation of this integral follows the same steps as the
evaluation of Ir in Lemma 2.
References
n
vk
Endnotes
ƒ
Batley R, Daly A (2006) On the equivalence between elimination-byaspects and generalized extreme value models of choice behavior.
J. Math. Psych. 50(5):456–467.
Batsell RR, Polking JC, Cramer R, Miller C (2003) Useful mathematical
relationships embedded in Tversky’s elimination by aspects model.
J. Math. Psych. 47(5–6):538–544.
Becker GM, DeGroot MH, Marschak J (1963) Stochastic models of choice
behavior. Behavioral Sci. 8(1):41–55.
Beggs S, Cardell S, Hausman J (1981) Assessing the potential demand
for electric cars. J. Econometrics 17(1):1–19.
Block HD, Marschak J (1960) Random orderings and stochastic theories
of responses. Olkin I, Ghurye S, Hoeffding W, Madow W, Mann H,
eds. Contributions to Probability and Statistics (Stanford University
Press, Stanford, CA), 97–132.
Chapman RG, Staelin R (1982) Exploiting rank ordered choice set data
within the stochastic utility model. J. Marketing Res. 19(3):288–301.
Chipman JS (1960) Stochastic choice and subjective probability. Willner D,
ed. Decisions, Values, and Groups, Vol. 1 (Pergamon Press, New York),
70–95.
Coombs CH (1958) On the use of inconsistency of preferences in psychological measurement. J. Experiment. Psych. 55(1):1–7.
Corbin R, Marley AAJ (1974) Random utility models with equality: An
apparent, but not actual, generalization of random utility models.
J. Math. Psych. 11(3):274–293.
Debreu G (1960) Review of R. D. Luce, individual choice behavior: A theoretical analysis. Amer. Econom. Rev. 50:186–188.
Fader P, McAlister L (1990) An elimination by aspects model of consumer
response to promotion calibrated on UPC scanner data. J. Marketing
Res. 27(3):322–332.
Gilbride T, Allenby G (2006) Estimating heterogeneous EBA and economic screening rule choice models. Marketing Sci. 25(5):494–509.
Görür D, Jäkel F, Rasmussen CE (2006) A choice model with infinitely
many latent features. Proc. 23rd Internat. Conf. Machine Learn.
Pittsburgh, PA (ACM, New York), 361–368.
Kohli R, Jedidi K (2007) Representation and inference of lexicographic
preference models and their variants. Marketing Sci. 26(3):380–399.
Krantz DH (1967) Rational distance function for multidimensional scaling. J. Math. Psych. 4(2):226–245.
Manrai AK, Sinha PK (1989) Elimination-by-cutoffs. Marketing Sci.
8(2):133–152.
McFadden D (1981) Econometric models of probabilistic choice. Manski C, McFadden D, eds. Structural Analysis of Discrete Data with
Econometric Applications (MIT Press, Cambridge, MA), 198–272.
Park SJ, Choi S (2013) A theoretical note on the number of free parameters in the elimination-by-aspects model. J. Math. Psych. 57(5):
255–259.
Pihlens DA (2008) The multi-attribute eliminations by aspects (MEBA)
model. Doctoral dissertation, University of Technology Sydney,
Sydney, Australia.
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Tversky A (1972a) Elimination by aspects: A theory of choice. Psych.
Rev. 79(4):281–299.
Tversky A (1972b) Choice by elimination. J. Math. Psych. 9(4):341–367.
Tversky A, Russo JE (1969) Similarity and substitutability in binary
choices. J. Math. Psych. 6:1–12.
Tversky A, Sattah S (1979) Preference trees. Psych. Rev. 86(6):542–573.
Rajeev Kohli is the Ira Leon Rennert Professor of Business,
Columbia University. His research interests include choice models, new product development, discrete optimization, and analysis
of algorithms. His previous research in the Journal of Marketing
Operations Research 63(3), pp. 512–526, © 2015 INFORMS
Research was the finalist for the Paul Green Award in 2008 and
the O’Dell Award in 2000.
Kamel Jedidi is the John A. Howard Professor of Business,
Columbia University. His research interests include pricing, diffusion of innovations, and the long-term impact of advertising and
promotions. He received Best Article Awards from International
Journal of Research in Marketing in 1998 and the Marketing Science Institute in 2000 and was finalist for 2009 Paul Green Award
for the Journal of Marketing Research and for the 2009 Longterm Impact Paper Award for Marketing/Management Science.